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While trying assignment problems of complex analysis I am unable to solve this particular question and so I am posting it here.

Let $f$ be a holomorphic function on $0<|z|<\epsilon$ , $\epsilon >0$ given by convergent Laurent series $\sum_{n=-\infty}^{\infty} a_{n} z^{n}$ . Given also that $\lim_{z\to0} |f(z)|= \infty$ .

Then which one is true.

  1. $a_{-1} \neq 0 $ and $a_{-n} =0$ for all $n\ge 2$.

  2. $a_{-N} \neq 0 $ for some $N>1$ and $a_{-n}=0$ for all $n >N$ .

  3. $a_{-n} =0$ for all $n>0$.

  4. $a_{-n} \neq 0$ for all $n>0$.

Although I have studied Ch- Laurent series from text book Complex variables with applications by Ponnusamy and Silvermanbut I am completely clueless on how this particular problem can be approched.

Can anyone please shed some light.

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    $\begingroup$ Hint: what kind of singularity does $f$ have at $0$? Removable, pole or essential? $\endgroup$ Commented Jul 31, 2020 at 18:23
  • $\begingroup$ $z^2/f(z)$ is holomorphic on a small disk $\endgroup$
    – reuns
    Commented Jul 31, 2020 at 21:19
  • $\begingroup$ @Robert Israel I think it's essential singularity. But can you please ellaborate on how it might be useful? $\endgroup$
    – user775699
    Commented Aug 4, 2020 at 8:17
  • $\begingroup$ No, $|f|$ can't go to $\infty$ at an essential singularity (see the Casorati-Weierstrass theorem). $\endgroup$ Commented Aug 4, 2020 at 14:15
  • $\begingroup$ @Robert Israel Statement of Casorati Weirestrauss theorem is that : If f(z) has an essential singularity at z=$z_{0}$ , then f(z) comes arbitrary close to every complex value in each deleted neighborhood of $z_{0}$ . So, I think it can approach $\infty$ . $\endgroup$
    – user775699
    Commented Aug 5, 2020 at 8:17

1 Answer 1

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Intuitively f(z)=1/z which satisfy given conditions in this way options (3) and (4) are wrong.If you consider f(z)=1/z² this will discard option (1). Hence, option (2) is correct.

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